Question

What volume (to the nearest 0.1 mL) of 6.80-M NaOH must be added to 0.250 L of 0.300-M HNO₂ to prepare a pH = 3.40 buffer?

Answer 1

pH = 3.40

pKa of HNO2 = 3.25

moles of HNO2 = 0.250 x 0.3 = 0.075

moles of NaOH = 6.80 V

HNO2 +    NaOH   ----------------> NaNO2 + H2O

0.075      6.80V                                0             0

0.075-6.80V        0                                  6.80 V

pH = pKa + log [salt / acid]

3.40 = 3.25 + log [6.80 V / 0.075-6.80V]

[6.80 V / 0.075-6.80V] = 1.4125

V = 6.46 x 10^-3 L

volume of NaOH = 6.46 mL